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ADVANCED PROBLEMS AND SOLUTIONS
dited by
RAYMOND E. WHITNEY
Lock Haven State College, Lock Haven, PA 17745
Send all comm unications concerning ADVANCED PROBLEMS AND SOLUTIONS to
RAYMOND E. WHITNEY, MATHEMATICS DEPARTMENT, LOCK HAVEN STATE C OLLEGE, LOCK
HAVEN, P-A 17745, This department especially welcomes problems believed to
be new or extending old results. Proposers should submit solutions or other
information that will assist the editor. To facilitate their consideration,
solutions should be submitted on separate signed sheets within two months of
publication of the problems,
H 307
Proposed by Larry Taylor, Briarwood, NY
(A) If p = 1 (mod 10) is prime, x =
/B ,
and a = . 7 (modp ,prove
that
a,
a + 1, a + 2, a + 3, and
a +
4 have the same quadratic character mod-
ulo p if and only if 11 < p = 1 or 11 (mod 60) and
(-2x/p)
= 1.
(B) If p = 1 (mod
60), (2x/p)
= 1, and
b
= ~
2
^ J ^
5 )
(mod
p ) ,
then
b,
b
+ 2,
b
+ 3, and
b +
4 have the same quadratic character modulo p. Prove
that (llab/p) = 1 .
H 3O8
Proposed by Paul Bruckman , Concord, CA
K ?n
( a
i'
a
2>
U
n)
L e t
[a
, a , . . . ,
a
n
]
= 7 - r d e n o t e t h e n t h c o n v e r -
g e n t o f t h e i n f i n i t e s i m p l e c o n t i n u e d f r a c t i o n
[a
l9
a
2
,
. . . ] ,
n
= 1 , 2 , . . . .
A l s o , d e f i n e p
0
= 1, ^
0
= 0 . F u r t h e r , d e f i n e
(1 )
W
n>k
= p
n
( a
x
, a
2
, . . . , a
n
) q
f e
( a
1 5
a
2
, . . . , a
fc
)
-
p
k
( a
x
, a
2
, . . . , ^
f e
) ^
n
t o
1 5
a
z
*
a
^ )
P
n
^ P
k
0 < f c < n .
Find a general formula for W
n>
^.
H 309
Proposed by David Singmaster, Polytechnic of the South Bank, London,
England
Let f be apermutation of
{l,
2, ..., m- 1} such that the terms i+f(i)
are all distinct (mod m). Characterize and/or enumerate such /. [Each such
/ gives a decomposition of the
m( m
+ 1) m-nomial coefficients, which are the
nearest neighbors of a given m-nomial coefficient, into msets of m+ 1 coef-
ficients which have equal products and are congruent by rotationsee Hoggatt
& Alexanderson, A Property of Multinomial Coeffieients,
The Fibonacci Quar-
terly 9, No. 4 (1971):351-356, 420-421.]
I have run a simple program to generate and enumerate such /, but can
see no pattern. The number Nof such permutations is given below for m
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Dec.1979
ADVANCED PROBLEMSANDSOLUTIONS
375
m 2 3 4 5 6 7 8 9 10
N
1 1 2 3 8 19 64 225 928
H-310 Proposed by V. E. Hoggatt, Jr., San Jose State University, San Jose, CA
Leta = (1 +
/5)/2
5
[na] =
a
n9
and
[na
2
] = b
n
.
Clearly,
a
n
+ n
=
b
n
.
a) Show that
if n=F
2 m + 1
,
then
a
n
= F
2m +2
and
n
=
F
2m+S
.
b) Show thatif n=F
2 m
, thena
n
= F
2m +1
*
anc
* ^
=
^im
i ~*
e) Show thatif n =-^2m
s
thena
n
= L
2m 1
and2?
n
=
^2m +2'
d) Show that if n =
2
m + l
5 t n e n
^ n
=
^ 2 ^+2 ~ *
a n d
^
=
^2m +
3
~
1 B
SOLUTIONS
EdAJtohslcit
Uotzt Starting with this issue, we shall indicate the issue and
date when each problem was proposed.
Continue
H-278 Proposed by V.E. Hoggatt, Jr., San Jose State University, San Jose, CA
(Vol. 16, No. 1, Feb. 1978)
Show J =
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376
ADVANCED PROBLEMSANDSOLUTIONS
[Dec
A Rare Mixture
H-279
Proposed by G. Wulczyn, Bucknell University, Lewisburg, PA
(Vol. 16, No. 1, Feb. 1978)
Establish the
F-L
identities:
(a) ^
n + 6 r
~ ^hr
+
* ^n
hr ~
n.
2r *
=
^
2vi 2> 6r^hn
12r
( b ) ^n+er 3
+
(k'tr
+ 2
*~
*'
^n+hr+2 ~ ^n 2r l^ ""
n
jp p ~p jp
2r
l
r
hr
2 Sr+ 3 hn
12r+B
Solution
by
Paul Bruckman, Concord,
CA
LmmcL 1: L
3m
- (-l)
m
L
m
=
5F
m
F
2m
.
Vtiooji L
3m
- (-l)
m
L
m
= a
3m
+ b
3m
- (ab)
m
(a
m
+ b
m
)
=
(a
m
-
b
m
)(a
2m
-b
2m
)
= 5F
m
F
2m
.
Imma 2: 5(F* - * * ) = F
U
_
V
F
U+ V
(L
U
_
V
L
U + V
-
4 ( - l )
w
) .
VKooji 25F
h
u
= (a
u
-b
u
)
h
=a
hu
+b
hu
- h(-l)
u
(a
2u
+b
2u
) + 6 .
T h e r e f o r e ,
25
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ADVANCED PROBLEMS
AND
SOLUTIONS
377
~ 5
F
2n +3m
L
Zn+3m(
F
3m
L
3m ~
^
_ 1
^
F
3m
L
m)
( a p p l y i n g L e m m a 3)
=
J
F
3m
F
kn
+
6
m
L
3m
"
( _ 1 ) L
*)
=
F
m
F
Zm
F
Bm
F
hn
+
6m ^
Lemma
^
T h e r e f o r e :
p^
_
((-l}
m
L
+ l)(F
k
- F
h
)-F
h
=FFFF
S e t t i n g
m - 2v
and
m - 2v
+ 1 y i e l d s ( a ) a n d ( b ) , r e s p e c t i v e l y .
Also solved
by the
proposer.
o
rn
H-280
Proposed
by P.
Bruckman , Concord, CA
(Vol. 16, No. 1, Feb. 1978)
P r o v e
t h e
c o n g r u e n c e s
(1) F
3
.
2
=
2*
+ 2
(mod 2
n + 3
) ;
(2 )
L
3
,
zn
= 2 + 2
2
+ 2
(mod2
2 n +1(
)
, n = 1, 2 , 3
Solution
by
Gregory Wulczyn, Bucknell University, Lewisburg,
PA
(1) n
=
1, F
e
= 8 = 8
(mod 16).
S i n c e L\
k
- 5F\
k
= 4 ,
(L
3k
,F
3k
)
= 2 ,
2
r
\F
3
,
2
n
9
r > 1
9
n >_ 1;
2
t
\L
3
.
z
n
i f ando n l y i f t = 1.
Assume F
3
.
2
n = 2
n+2
(mod 2
3
) =
2
n + 2
(mod 2
n + lf
)
F
3
.
2
n
+
i
= F
3
.
2
n L
3
.
2
n
E
2
n + 3
(mod 2
n + I t
.
(2)
n =
1, L
6
=
18
E 2 +2
h
(mod2
6
)
,
L
2
2k
=
L ^
+
2.
Assume L
3
.
2
-=
2 +
2
2 n
(mod2
2 n f
)
L
3
.
2
n
+
i
=L
2
.
2
- 2 = 2 +2
2 n
>
+
2
4 n+f
(mod2
4 n 5
)
3
.
2
+i E 2 +
2
2 n
f
(mod
2
2 n 6
9
n >1.
Also
solved by the proposer, who noted that this is Corollary 6 in Periodic
Continued Fraction Representations of Fibonacci-type Irrationals / by V. E.
Hoggatt,
Jr.
& Paul
S.
Bruckman,
in The
Fibonacci Quarterly
15,
No .
3
(1977):
225-230.